(cosA)^2*sinA的最大值怎么求呀?
人气:256 ℃ 时间:2020-05-13 21:17:21
解答
高考复习啊,那就用导数:y = (cosA)^2*sinA= [1 - (sinA)^2]*sinA= sinA - (sinA)^3y' = cosA - 3(sinA)^2*cosA= cosA[1 - 3*(sinA)^2]= 0cosA = 0 .y = 0(sinA)^2 = 1/3sinA = ±√3/3(cosA)^2 = 2/3y = ±2√3/9 ....
推荐
猜你喜欢